Can we say that $S^{\infty} / x \sim -x \approx \Bbb RP^{\infty}\ $?
I know that $\Bbb RP^n = S^n/x \sim -x.$ I am wondering on whether the similar thing holds for $\Bbb RP^{\infty}.$ I am trying by taking a map $q : S^{\infty} \longrightarrow \Bbb R P^{\infty}$ defined by $x \mapsto \{x,-x\}.$ Clearly this map is surjective. Now we need to show that this map is continuous. Since $\Bbb RP^{\infty}$ is endowed with the weak topology and quotient topology is the strongest topology with respect to which the quotient map is continuous it follows that $q$ is continuous. So we get a continuous surjective map from $S^{\infty}$ onto $\Bbb RP^{\infty}$ which identifies antipodal points. Hence by universal property of quotient topology it induces a bijective continuous map $\overline {q} : S^{\infty}/x \sim -x \longrightarrow \Bbb RP^{\infty}.$ So in order to show that $\overline {q}$ is a homeomorphism we need to show that $q$ is a quotient map. Let $U \subseteq \Bbb RP^{\infty}$ be such that $q^{-1} [U]$ is open in $S^{\infty}/x \sim -x.$ Since $S^{\infty}$ is endowed with the weak topology it follows that $q^{-1}[U] \cap S^n$ is open in $S^n.$ Now I think $q^{-1} [U] \cap S^n = q^{-1} [U \cap \Bbb RP^n].$ Now since $q \rvert_{\Bbb S^n} : S^n \longrightarrow \Bbb RP^n$ is the usual quotient map and $q^{-1} [U \cap \Bbb RP^n] = {(q \rvert_{S^n})}^{-1} [U \cap \Bbb RP^n]$ it follows that $U \cap \Bbb R P^n$ is open in $\Bbb RP^n$ and since $\Bbb R P^{\infty}$ is endowed with the weak topology it follows that $U$ is open in $\Bbb R P^{\infty}$ and hence $q$ is indeed a quotient map. This completes the proof.
Does the above reasoning make sense? Would anybody kindly check it?
Thanks in advance.